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I have never seen a real-analytic approach before to evaluate integrals of the form below $$\int_a^b\text{elementary function}(x)\,dx=\text{constant involving}\,W(\cdot)\,\text{in its simplest form}\tag1.$$

For instance, on MSE, all use the residue theorem:

And the same applies to some of the wider literature I have come across:

So, my question is this:

Does anyone know of a proof of an identity of the form in $(1)$ that involves only real analysis (i.e. does not assume the existence of $\sqrt{-1}$)?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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  • Does $\int\limits_{\infty }^1 e^{1-x},dx=W\left(-\frac{1}{e}\right)=-1$ count? – Steven Clark Jul 28 '22 at 16:33
  • @StevenClark No, which is the reason I added "in its simplest form" (or "in most simplified form" before this edit) so that we don't have trivial cases like $\int_0^12x,dx=1=W(10)/W(10)$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jul 28 '22 at 16:35
  • My example is less trivial than your example, and is $\int\limits_{\infty }^1 e^{1-x},dx$ not a valid integral representation of $W\left(-\frac{1}{e}\right)$? – Steven Clark Jul 28 '22 at 16:57
  • @StevenClark I think there is a bit of misunderstanding -- I'm trying to find a valid integral representation (elementary integrand) for values of a constant involving $W(\cdot)$ such that it cannot be simplified to a constant that doesn't involve $W$. So $1/(1+W(2))$ would be valid, but constants like $W(-1/e)=-1$ or $\sin W(3e^3)^2=\sin9$ would not. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jul 28 '22 at 17:01
  • Does one also have to prove a result involving $W(x)$ has no closed form that does't involve the $W(x)$ function? – Steven Clark Jul 28 '22 at 17:09
  • @StevenClark No, I think as long as your $x$ does not obviously simplify to $ye^y$ for some elementary $y$ then it's accepted (like obvious manipulations of $W((\sqrt2+1)e^{\sqrt2+1})$. Since otherwise we'd be running into much deeper holes involving Schanuel's conjecture. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jul 28 '22 at 17:22
  • Do you consider numerical integration a real-analytic approach? Some of your examples can be evaluated via numerical integration, but some of your examples also simplify which seems inconsistent with your desire to find an example that doesn't simplify. Excluding results that simplify seems rather arbitrary assuming one can show the example is inherently related to the Lambert W function. Trivial cases such as $\frac{W(10)}{W(10)}$ don't generally meet this criteria, but it seems to me $\int\limits_{-\infty}^{\infty}{e^x+1\over (e^x-x+1)^2+\pi^2}\mathrm dx=1$ does meet this criteria. – Steven Clark Aug 06 '22 at 16:54
  • @StevenClark No (I'm looking for an analytical proof rather than numerical). In that example, I'm actually not referring to the end result $\int_{-\infty}^{\infty}{e^x+1\over (e^x-x+1)^2+\pi^2},dx=1$ but rather the lemma $\int_{-\infty}^{\infty}\frac{a^2,dx}{(e^x-ax-b)^2+(a\pi)^2}=\frac{1}{1+W\left(\frac{1}{a}e^{-b/a}\right)}$ in Jack's answer, which he said was evaluated only through the residue theorem. To qualify as an answer here, an example would be to prove the latter identity using real analysis when, say, $a=2$ and $b=3$, since $W(e^{-3/2}/2)$ doesn't have an obvious simplification. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Aug 06 '22 at 16:57
  • @TheSimpliFire Each thread you linked used either contour integration, residue theorem, Cauchy integral theorem, or another complex analysis theorem. Is a solution without these, but still using complex numbers, like in transforms or summation formulas, ok? – Тyma Gaidash Nov 12 '23 at 21:13
  • @ТymaGaidash No, the proof must not use the fact that $\sqrt{-1}$ exists. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 13 '23 at 22:50

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